"pmb"  wrote in message 
news:rqidnRoyYrM8aiPbnZ2dnUVZ8qKvnZ2d@bt.com...
> It's a long time since I was at school, and I've forgotten a lot. But I'm 
> trying to solve three equations. They look like they're solvable, but I 
> can't do it. It's possible that they contain some absurdity that makes 
> them unsolvable. But here they are:
>
> a=10+b+c
> 2b=10+a+c
> 3c=10+a+b
>
> Is there as solution?


There is no reason why a, b, and/or c could not be -ve.

There are several approachs but the golden rule is to have one equation per 
unknown ie you need 1 equation to find 1 unknown, 2 for 2, etc.
Algebraic methods generally rely on this.

Here you've 3 of each so, in essence, if there is a solution you should be 
able to find it. I'd start by reducing the problem to 2 unknowns, use the 
first equation to subsitute for a in the other 2.

Therefore:

2b = 10 + 10+b+c   +c      or    2b = 20 + b + 2c   or     b = 20+2c

3c = 10 + 10+b+c +b        or    3c = 20 + 2b +c    or   2c= 20 + 2b  or 
c = 10 +b

You've now got 2 equations with 2 unknowns

b = 20+2c
c=  10+b

Repeat the process, reducing to 1 equ and 1 unknown

Use the first equ to subsitute in the second for b

c = 10 + 20 + 2c

-30 = c

Now back track:

b = 20 + 2c   or    b = 20 + 2(-30)  so b = -40

You've now got b and c and use  on of your original equations to find a

a = 10 + b + c   so a = -60

So a = -60, b= -40, c = -30

Good practice is to check in all of your original equations, just in case:

(1)       a=10+b+c                         -60 = 10 -40 - 30     OK
(2)      2b=10+a+c                        -80 = 10 -60- 30      OK
(3)      3c=10+a+b                        -90 = 10 -60 -40       OK



-- 
73
Brian, G8OSN
www.g8osn.org.uk

Now your amateur licence is free, why not send at least £15 per year to 
support the
Radio Communications Foundation or STELAR?